codeforces--707E. Garlands（二维树状数组）

cf 707E

题解

参考题解

#include <bits/stdc++.h>using namespace std;const double eps = 1E-8;const int dx[4] = {1, 0, 0, -1};const int dy[4] = {0, -1, 1, 0};const int inf = 0x3f3f3f3f;const int N = 1E5 + 7;#define MS(a, x) memset(a, x, sizeof(a))#define MP make_pair#define fst first#define sec second#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1typedef long long LL;const int maxn = 2000 + 3;// AC info: 1482 ms 72900 KBstruct Cell{    int x, y;    LL val;};int n, m, k, q;vector<Cell> garland[maxn];LL c[maxn][maxn];bool on[maxn], last[maxn];int lowbit(int i) { return i & (-i); }void add(int x, int y, int v){    for(int i = x; i <= n; i += lowbit(i)){        for(int j = y; j <= m; j += lowbit(j)) c[i][j] += v;    }}LL sum(int x, int y){    LL s = 0;    for(int i = x; i ; i -= lowbit(i)){        for(int j = y; j ; j -= lowbit(j)) s += c[i][j];    }    return s;}int main(){#ifdef LOCALfreopen("data.in", "r", stdin);#endif // LOCAL    scanf("%d %d %d", &n, &m, &k);    for(int i = 1; i <= k; ++i){        int lights;        scanf("%d", &lights);        for(int j = 0; j < lights; ++j){            int x, y, val;            scanf("%d %d %d", &x, &y, &val);            garland[i].push_back(Cell{x, y, val});        }        // turn on        on[i] = true;    }    scanf("%d", &q);    char op[8];    int x1, y1, x2, y2;    for(int cnt = 0; cnt < q; ++cnt){        scanf("%s", op);        if(op[0] == 'S'){            int which;            scanf("%d", &which);            on[which] ^= 1;        }        else{            scanf("%d %d %d %d", &x1, &y1, &x2, &y2);            // 每一次ASK，都要计算每条链的贡献            for(int i = 1; i <= k; ++i){                // 如果和上一次的状态不同，更新                if(last[i] != on[i]){                    for(int j = 0; j < garland[i].size(); ++j){                        // 第i条链的第j个灯                        if(on[i]) add(garland[i][j].x, garland[i][j].y, garland[i][j].val);                        else add(garland[i][j].x, garland[i][j].y, -garland[i][j].val);                    }                }                // 记录上一次的状态                last[i] = on[i];            }            LL ans = sum(x2, y2) - sum(x2, y1 - 1) - sum(x1 - 1, y2) + sum(x1 - 1, y1 - 1);            printf("%I64d\n", ans);        }    }    return 0;}